3.3.3 Bohr-Sommerfeld Quantization#

Worked solutions for the homework problems in the 3.3.3 Bohr-Sommerfeld Quantization lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Harmonic oscillator. For the harmonic oscillator \(V(x) = \frac{1}{2}m\omega^{2}x^{2}\), the classical turning points at energy \(E\) are \(x = \pm a\) with \(a = \sqrt{2E/(m\omega^{2})}\).

(a) Compute the orbit integral \(\oint p\,\mathrm{d}x = 2\int_{-a}^{a}\sqrt{2m(E - \tfrac{1}{2}m\omega^{2}x^{2})}\,\mathrm{d}x\) and show that it equals \(2\pi E/\omega\).

(b) Apply the Bohr-Sommerfeld rule \(\oint p(x)\,\mathrm{d}x = 2\pi\hbar(n + \tfrac{1}{2})\) to derive \(E_{n} = \hbar\omega(n + 1/2)\). Note that this is the exact quantum result.

Solution.

(a) At the turning points \(E = \tfrac{1}{2}m\omega^{2}a^{2}\), so \(E\) and \(a\) are related by \(2E = m\omega^{2}a^{2}\). Use this to rewrite the momentum:

where the middle step used \(2mE = m^{2}\omega^{2}a^{2}\). The orbit integral runs once around the closed phase-space loop — out from \(-a\) to \(+a\) and back — which is twice the line integral between the turning points:

The remaining integral is the area under a semicircle of radius \(a\), \(\int_{-a}^{a}\sqrt{a^{2}-x^{2}}\,\mathrm{d}x = \tfrac{1}{2}\pi a^{2}\). Hence

Finally substitute \(a^{2} = 2E/(m\omega^{2})\):

(b) Apply the Bohr-Sommerfeld rule \(\oint p\,\mathrm{d}x = 2\pi\hbar(n + \tfrac{1}{2})\) at \(E = E_{n}\):

so \(E_{n} = \hbar\omega(n + \tfrac{1}{2})\) for \(n = 0, 1, 2, \ldots\)

This coincides with the exact quantum spectrum, zero-point energy included. The agreement is not luck: for a quadratic potential the higher-order (\(\hbar^{2}\) and beyond) corrections to the WKB phase vanish identically, so the leading semiclassical result is already exact. The Maslov \(\tfrac{1}{2}\) is precisely what supplies the ground-state energy \(\tfrac{1}{2}\hbar\omega\); without it one would predict a vanishing zero-point energy.

2. Particle in a box. A particle of mass \(m\) is confined between hard walls at \(x = 0\) and \(x = a\). Inside the box \(V = 0\), so \(p = \sqrt{2mE}\).

(a) For hard walls, the Maslov index is \(\mu = 1\) per wall (versus \(\mu = 1/2\) at a soft turning point), so use the generalized rule \(\oint p\,\mathrm{d}x = 2\pi\hbar\bigl(n + (\mu_a+\mu_b)/2\bigr)\) with \(\mu_{a} = \mu_{b} = 1\).

(b) Show that this gives \(E_{n} = (n+1)^{2}\pi^{2}\hbar^{2}/(2ma^{2})\) for \(n = 0, 1, 2,\ldots\), which is the exact spectrum (with the conventional relabeling \(n\to n-1\)).

(c) Explain physically why the hard-wall Maslov index differs from the soft case: what does the wave do at a rigid boundary versus at a smooth turning point?

Solution.

(a) Inside the box the potential is flat, so the momentum \(p = \sqrt{2mE}\) is a constant. The classical orbit is a particle bouncing wall to wall: it travels from \(x = 0\) to \(x = a\) and back, so

Both boundaries are hard walls, \(\mu_{a} = \mu_{b} = 1\), so \((\mu_{a}+\mu_{b})/2 = 1\) and the generalized rule \(\oint p\,\mathrm{d}x = 2\pi\hbar\bigl(n + (\mu_a+\mu_b)/2\bigr)\) reads

(b) Equate the two expressions for the orbit integral:

so \(\sqrt{2mE_{n}} = \pi\hbar(n+1)/a\).

Square both sides and solve for the energy:

so \(E_{n} = (n+1)^{2}\pi^{2}\hbar^{2}/(2ma^{2})\) for \(n = 0, 1, 2, \ldots\)

This is the exact infinite-square-well spectrum. With the conventional relabeling \(n \to n-1\) (so the ground state carries the label \(1\) rather than \(0\)), it takes the familiar form \(E_{n} = n^{2}\pi^{2}\hbar^{2}/(2ma^{2})\) for \(n = 1, 2, 3, \ldots\) The semiclassical rule is exact here because the only non-trivial physics is the boundary phase, which the Maslov bookkeeping captures exactly — the interior wavefunction is a pure plane wave with no curvature correction.

(c) At a rigid wall the potential jumps to infinity, so the wavefunction must vanish there, \(\psi = 0\). An incident wave reflects with a sign flip — a hard node is pinned at the boundary, exactly as for a wave on a string with a fixed end. The reflected wave is phase-shifted by a full \(\pi\), which is the content of \(\mu = 1\).

At a soft turning point the potential is finite and smooth; \(E = V\) there but \(\psi\) does not vanish. The wave does not reflect abruptly — it bends over and leaks into the classically forbidden region, decaying exponentially beyond the turning point (the Airy-function behaviour of §3.3.2). This gentle turnover costs only half the phase: a \(\pi/2\) shift, i.e. \(\mu = \tfrac{1}{2}\). The difference is precisely the difference between a wave slamming into an impenetrable barrier and a wave easing through a smooth classical boundary.

3. Linear potential. A particle in a half-line gravitational-like potential \(V(x) = Fx\) for \(x \geq 0\), with a hard wall at \(x = 0\).

(a) At energy \(E\), find the soft turning point \(x_{a} = E/F\) and compute \(\int_{0}^{x_{a}}\sqrt{2m(E - Fx)}\,\mathrm{d}x = \tfrac{(2mE)^{3/2}}{3mF}\).

(b) Write the Bohr-Sommerfeld rule with the correct Maslov indices (\(\mu = 1\) at the hard wall, \(\mu = 1/2\) at the soft turning point) and solve for \(E_{n}\).

(c) Show that the energy spacing \(\Delta E_{n}\) decreases with \(n\), unlike the harmonic oscillator. Why does a linear potential produce non-uniform level spacing?

Solution.

(a) At the soft turning point the particle is momentarily at rest, so all of its energy is potential: \(E = V(x_{a}) = F x_{a}\), hence

For the integral, substitute \(u = E - Fx\), so \(\mathrm{d}u = -F\,\mathrm{d}x\); the limits map \(x = 0 \to u = E\) and \(x = x_{a} \to u = 0\):

Written compactly, this is

(The two forms are identical: \((2mE)^{3/2} = 2m\sqrt{2m}\,E^{3/2}\), so dividing by \(3mF\) leaves \(\tfrac{2\sqrt{2m}}{3F}E^{3/2}\). The compact form \(\tfrac{(2mE)^{3/2}}{3mF}\) is the one quoted in the problem statement, and is the value used in parts (b) and (c).)

(b) The classical orbit is a particle launched up the ramp, decelerating to rest at \(x_{a}\), and falling back to bounce off the hard wall at \(x = 0\). The orbit integral covers the up-and-back trip:

The two boundaries are different: a hard wall at \(x = 0\) (\(\mu_{a} = 1\)) and a soft turning point at \(x_{a}\) (\(\mu_{b} = \tfrac{1}{2}\)). The Maslov term is \((\mu_{a}+\mu_{b})/2 = (1 + \tfrac{1}{2})/2 = \tfrac{3}{4}\), so the generalized rule \(\oint p\,\mathrm{d}x = 2\pi\hbar\bigl(n + (\mu_a+\mu_b)/2\bigr)\) is

Solve for \(E_{n}\):

so \(E_{n} = (9\pi^{2}\hbar^{2}F^{2}/(8m))^{1/3}(n + \tfrac{3}{4})^{2/3}\) for \(n = 0, 1, 2, \ldots\)

The energy levels scale as \(E_{n} \propto (n + 3/4)^{2/3}\) — the characteristic \(n^{2/3}\) growth of a linear potential.

(c) Because \(E_{n} \propto (n + 3/4)^{2/3}\) with exponent \(2/3 < 1\), the spectrum is concave in \(n\). The spacing is the discrete derivative; for large \(n\),

which decreases as \(n^{-1/3}\). The harmonic oscillator, by contrast, has \(E_{n} \propto n\) exactly, so its spacing \(\hbar\omega\) is constant.

The physical reason is the classical orbital period. In the semiclassical regime the level spacing equals \(\hbar\) times the classical orbital frequency, \(\Delta E_{n} \approx \hbar\,\omega_\text{cl}(E_{n})\) — indeed \(\mathrm{d}(\oint p\,\mathrm{d}x)/\mathrm{d}E\) is exactly the orbital period. For the harmonic oscillator the period is independent of amplitude (isochronism), so \(\omega_\text{cl}\) and hence \(\Delta E\) are energy-independent. For the linear ramp the turning point \(x_{a} = E/F\) recedes as the energy grows, the orbit gets longer, and the period lengthens (one finds \(\omega_\text{cl} \propto E^{-1/2}\)). A slower orbit means a smaller energy quantum, so the levels crowd together at high \(n\).

4. Anharmonic oscillator. Consider \(V(x) = \tfrac{1}{2}m\omega^{2}x^{2} + \lambda x^{4}\) with \(\lambda > 0\) small.

(a) Explain qualitatively why the quartic term steepens the potential at large \(\vert x\vert\), shrinking the classical turning points compared to the pure harmonic case at the same energy.

(b) Argue that the orbit integral \(\oint p\,\mathrm{d}x\) is therefore smaller than for the harmonic oscillator at the same energy, so Bohr-Sommerfeld predicts higher energy levels. Are the energy spacings larger or smaller than \(\hbar\omega\)?

(c) For which regime (low \(n\) or high \(n\)) is the quartic correction most important? Use the ratio \(\lambda x^{4}/(\tfrac{1}{2}m\omega^{2}x^{2})\) at the turning point.

Solution.

(a) The added term \(\lambda x^{4}\) is strictly positive for \(x \neq 0\) (since \(\lambda > 0\)), and it grows faster than the quadratic part: the ratio \(\lambda x^{4}/(\tfrac{1}{2}m\omega^{2}x^{2}) = 2\lambda x^{2}/(m\omega^{2})\) increases without bound as \(\vert x\vert\) grows. So at large displacement the quartic well sits above the pure harmonic well. The classical turning point is where \(V(x_{t}) = E\); since the anharmonic potential reaches the value \(E\) sooner — at a smaller \(\vert x\vert\) — its turning points \(\pm x_{t}\) lie inside the harmonic turning points \(\pm a\) for the same energy \(E\). The quartic wall “catches” the particle earlier.

(b) The orbit integral \(\oint p\,\mathrm{d}x = 2\int_{-x_{t}}^{x_{t}}\sqrt{2m(E - V)}\,\mathrm{d}x\) is reduced by the quartic term in two reinforcing ways: the integration range \([-x_{t}, x_{t}]\) is narrower (part (a)), and at every interior point the potential is higher, so \(E - V\) is smaller and the integrand \(\sqrt{2m(E-V)}\) is smaller. Both effects shrink the integral, so at a fixed energy \(E\)

Now read the Bohr-Sommerfeld rule the other way around. The right-hand side \(2\pi\hbar(n+\tfrac{1}{2})\) is fixed by the quantum number \(n\). Since the anharmonic orbit integral accumulates more slowly with energy, a larger \(E\) is needed to reach the same value \(2\pi\hbar(n+\tfrac{1}{2})\). Hence \(E_{n}^\text{anh} > \hbar\omega(n+\tfrac{1}{2})\) — every level is pushed up.

For the spacing, recall \(\mathrm{d}(\oint p\,\mathrm{d}x)/\mathrm{d}E\) is the classical period \(T(E)\), so \(\Delta E_{n} \approx 2\pi\hbar/T(E_{n})\). The quartic term stiffens the restoring force (\(-V' = -m\omega^{2}x - 4\lambda x^{3}\) is stronger than harmonic), so the oscillation is faster: \(T < 2\pi/\omega\) and shrinks further as the amplitude grows. Therefore the spacings are larger than \(\hbar\omega\), and they increase with \(n\). A harder-than-quadratic well always has a widening spectrum (the extreme case being the box, \(E_{n}\propto n^{2}\)).

(c) The relative size of the quartic correction is measured at the turning point, where the displacement is largest:

To leading order the turning point satisfies \(\tfrac{1}{2}m\omega^{2}x_{t}^{2} \approx E\), i.e. \(x_{t}^{2} \approx 2E/(m\omega^{2})\), so the ratio becomes

The ratio grows with the energy and hence with the quantum number. The quartic correction is therefore negligible at low \(n\) — near the bottom of the well the amplitude is tiny and the potential is effectively harmonic — and most important at high \(n\), where the large-amplitude orbit explores the steep quartic walls. This is why anharmonic corrections to molecular vibrational spectra show up as a progressive widening (or, for softer wells, narrowing) of the level spacing at higher excitation.

5. Two-dimensional anisotropic oscillator. A particle moves in \(V(x,y) = \tfrac{1}{2}m\omega_{x}^{2}x^{2} + \tfrac{1}{2}m\omega_{y}^{2}y^{2}\) with \(\omega_{x} \neq \omega_{y}\).

(a) The motion separates: \(x\) and \(y\) each oscillate independently. Apply Bohr-Sommerfeld to each direction to derive \(E_{n_{x},n_{y}} = \hbar\omega_{x}(n_{x} + \tfrac{1}{2}) + \hbar\omega_{y}(n_{y} + \tfrac{1}{2})\).

(b) For \(\omega_{x} = 2\omega_{y} = 2\omega\), list the first five distinct energy levels in units of \(\hbar\omega\) and identify any degeneracies.

(c) The same construction fails for non-separable (in particular chaotic) potentials. In one sentence, explain why the rule “quantize each closed action variable independently” requires the system to be integrable (as many conserved quantities as degrees of freedom).

Solution.

(a) The potential is a sum of an \(x\)-part and a \(y\)-part, and the kinetic energy splits the same way, \(H = H_{x} + H_{y}\) with \(H_{x} = p_{x}^{2}/2m + \tfrac{1}{2}m\omega_{x}^{2}x^{2}\) and likewise for \(y\). The two coordinates do not couple, so the total energy partitions as \(E = E_{x} + E_{y}\), and each coordinate executes an independent 1D harmonic oscillation with its own conserved energy. Bohr-Sommerfeld applies separately to each closed loop in the \((x,p_{x})\) and \((y,p_{y})\) planes. Each is a 1D harmonic oscillator with two soft turning points, so by Problem 1,

Adding the two independently quantized pieces,

(b) With \(\omega_{x} = 2\omega\) and \(\omega_{y} = \omega\),

So the energy in units of \(\hbar\omega\) is \(2n_{x} + n_{y} + \tfrac{3}{2}\). Enumerating \((n_{x}, n_{y})\):

The first five distinct levels are \(\tfrac{3}{2}, \tfrac{5}{2}, \tfrac{7}{2}, \tfrac{9}{2}, \tfrac{11}{2}\) (in units of \(\hbar\omega\)). Degeneracies appear from the third level upward: distinct \((n_{x},n_{y})\) pairs give the same energy because \(2n_{x} + n_{y}\) can be reached in several ways once it is \(\geq 2\). This is a consequence of the commensurate (rational-ratio) frequencies \(\omega_{x}:\omega_{y} = 2:1\); for an irrational ratio every level would be non-degenerate.

(c) Quantizing “each closed action variable independently” presupposes that the phase-space motion winds around invariant tori on which each action \(\oint p_{i}\,\mathrm{d}q_{i}\) is separately well-defined and conserved — and that requires the system to be integrable, i.e. to possess as many independent conserved quantities as degrees of freedom; a chaotic (non-integrable) system has too few constants of motion, its trajectories fill phase-space regions ergodically rather than lying on tori, and no global action variables exist to quantize.

6. Hydrogen atom (circular orbits). For the Coulomb potential \(V(r) = -q_{e}^{2}/(4\pi\epsilon_{0}r)\), consider circular orbits with angular momentum \(L = n_{\varphi}\hbar\).

(a) From the force balance \(mv^{2}/r = q_{e}^{2}/(4\pi\epsilon_{0}r^{2})\) and \(L = mvr = n_{\varphi}\hbar\), show that \(r_{n_{\varphi}} = n_{\varphi}^{2}a_{0}\) where \(a_{0} = 4\pi\epsilon_{0}\hbar^{2}/(m q_{e}^{2})\) is the Bohr radius.

(b) Use the virial theorem (\(E = -T = V/2\) for the Coulomb potential) to derive \(E_{n_{\varphi}} = -13.6\,\text{eV}/n_{\varphi}^{2}\).

(c) Adding the radial Bohr-Sommerfeld quantization with a radial quantum number \(n_{r}\) promotes \(n_{\varphi}\) to \(n = n_{r} + n_{\varphi}\). Explain why this reproduces the same formula \(E_{n} = -13.6\,\text{eV}/n^{2}\) with additional degeneracy.

Solution.

(a) The angular-momentum quantization \(L = mvr = n_{\varphi}\hbar\) fixes the speed in terms of the radius, \(v = n_{\varphi}\hbar/(mr)\). The force balance for a circular orbit (Coulomb attraction supplies the centripetal force) is

so \(mv^{2} = q_{e}^{2}/(4\pi\epsilon_{0}r)\).

Substitute \(v = n_{\varphi}\hbar/(mr)\) into the left-hand side:

so \(n_{\varphi}^{2}\hbar^{2}/(m\,r) = q_{e}^{2}/(4\pi\epsilon_{0})\).

Solving for \(r\),

The allowed circular radii grow as \(n_{\varphi}^{2}\), in units of the Bohr radius \(a_{0}\).

(b) For a circular orbit the kinetic energy is \(T = \tfrac{1}{2}mv^{2}\), and the force balance of part (a) gives \(mv^{2} = q_{e}^{2}/(4\pi\epsilon_{0}r)\), so

which is the virial relation for the Coulomb (\(\propto 1/r\)) potential. The total energy is then

The prefactor is the Rydberg energy. Substituting \(a_{0} = 4\pi\epsilon_{0}\hbar^{2}/(m q_{e}^{2})\),

so that

(c) Circular orbits are only the special case of zero radial motion. A general bound orbit is an ellipse on which \(r\) also oscillates between a perihelion \(r_{-}\) and an aphelion \(r_{+}\) — two soft turning points. The radial libration carries its own action, which must be separately quantized. To carry this out, use the conserved angular momentum \(L = n_{\varphi}\hbar\) to eliminate the angular degree of freedom from the energy. With \(k \equiv q_{e}^{2}/(4\pi\epsilon_{0})\),

For bound motion \(E < 0\); write \(\vert E\vert = -E > 0\). The turning points are the roots of \(p_{r} = 0\), equivalently \(-2m\vert E\vert r^{2} + 2mk\,r - L^{2} = 0\):

Factor the radicand using these roots,

so

The radial action is then

The remaining integral has the closed-form value (Sommerfeld 1916; verifiable by the substitution \(r = \tfrac{1}{2}(r_{+}+r_{-}) - \tfrac{1}{2}(r_{+}-r_{-})\cos\theta\), or by contour methods),

Substituting the values of \(r_{+} + r_{-}\) and \(r_{+}r_{-}\),

Now apply Bohr-Sommerfeld separately to the two periodic motions. The angular motion is a ring (no turning points, \(\mu_{\varphi} = 0\)), so \(\oint p_{\varphi}\,\mathrm{d}\varphi = 2\pi L = 2\pi n_{\varphi}\hbar\) — already used. The radial motion has two soft turning points; we take its quantization in the old-Sommerfeld form \(\oint p_{r}\,\mathrm{d}r = 2\pi n_{r}\hbar\) with \(n_{r} = 0,1,2,\ldots\) (see remark below on Maslov bookkeeping). Then

so \(k\sqrt{m/(2\vert E\vert)} = (n_{r} + n_{\varphi})\,\hbar \equiv n\,\hbar\).

The radial and angular quantum numbers enter the energy only through their sum \(n\). Squaring and solving,

This is the same prefactor as in part (b), but now \(n = n_{r} + n_{\varphi}\) plays the role of the principal quantum number.

Why energy depends only on the sum. The Coulomb \(1/r\) potential is the unique attractive central potential (together with the isotropic harmonic oscillator) whose bound orbits close — every classical orbit retraces itself after one period. The radial and angular motions therefore share a common period, and the energy is fixed by the semi-major axis of the Kepler ellipse alone, independent of eccentricity. Different splittings \((n_{r},n_{\varphi})\) at fixed \(n\) correspond to orbits of different eccentricities (nearly circular: \(n_{\varphi}\approx n\), \(n_{r}\approx 0\); thin elongated ellipse: \(n_{\varphi}\) small, \(n_{r}\) large), but they share the same semi-major axis and hence the same energy. Each level \(n\) therefore acquires a degeneracy \(n_{\varphi} = 1, 2, \ldots, n\) (with \(n_{r} = n - n_{\varphi} \ge 0\)). This semiclassical degeneracy is the shadow of the exact \(n^{2}\)-fold hydrogen degeneracy, ultimately traced to the hidden \(\mathrm{SO}(4)\) symmetry of the \(1/r\) potential.

Remark on Maslov bookkeeping. Two soft radial turning points would naively contribute a \(+\tfrac{1}{2}\) Maslov correction, giving \(\oint p_{r}\,\mathrm{d}r = 2\pi\hbar(n_{r} + \tfrac{1}{2})\). For 3D radial WKB, however, an additional Langer correction shifts the effective angular momentum entering \(p_{r}\) from \(L = n_{\varphi}\hbar\) to \(L_\text{eff} = (n_{\varphi} + \tfrac{1}{2})\hbar\). The two half-integer shifts cancel in the combination \(k\sqrt{m/(2\vert E\vert)} = (n_{r}+\tfrac{1}{2}) + (n_{\varphi}+\tfrac{1}{2})\hbar = (n_{r}+n_{\varphi}+1)\hbar\), restoring the integer principal quantum number \(n = n_{r}+n_{\varphi}+1 \ge 1\) and the same \(E_{n}\). The old-Sommerfeld bookkeeping used above is the cleanest route; either accounting reproduces the exact hydrogen spectrum, a coincidence special to the \(1/r\) potential.

7. Correspondence principle. In the classical limit (\(n \gg 1\)), the energy spacing \(\Delta E = E_{n+1} - E_{n}\) should equal \(\hbar\omega_\text{cl}\), where \(\omega_\text{cl}\) is the classical orbital frequency.

(a) For the harmonic oscillator, verify \(\Delta E = \hbar\omega\) for all \(n\).

(b) For a power-law potential \(V(x) = C\vert x\vert^{\alpha}\), Bohr-Sommerfeld gives \(E_{n} \propto n^{2\alpha/(\alpha+2)}\). Show that \(\Delta E/E_{n}\to 0\) as \(n\to\infty\), so the spectrum becomes quasi-continuous in the classical limit.

(c) For hydrogen (\(E_{n}\propto -1/n^{2}\)), compute the classical orbital frequency \(\omega_\text{cl} = \Delta E/\hbar\) at large \(n\) and show it scales as \(n^{-3}\). This is the radiation frequency in transitions between adjacent high-\(n\) Rydberg levels.

Solution.

(a) From Problem 1, \(E_{n} = \hbar\omega(n + \tfrac{1}{2})\), so

The spacing is exactly \(\hbar\omega\) for every \(n\), not merely asymptotically. This matches the classical orbital frequency: a harmonic oscillator is isochronous, oscillating at \(\omega\) regardless of amplitude, so \(\omega_\text{cl} = \omega\) at all energies and \(\Delta E = \hbar\omega_\text{cl}\) holds for the harmonic oscillator down to the ground state.

(b) Write \(E_{n} = A\,n^{\beta}\) with \(\beta = 2\alpha/(\alpha+2)\). Since \(\alpha > 0\), the exponent satisfies \(0 < \beta < 2\); in particular \(\beta < 2\), and for any potential softer than the box \(\beta\) is finite. (Sketch of the scaling: the turning point obeys \(C x_{t}^{\alpha} = E\), so \(x_{t}\propto E^{1/\alpha}\); rescaling \(x = x_{t}u\) in \(\oint p\,\mathrm{d}x = 4\int_{0}^{x_{t}}\sqrt{2m(E - Cx^{\alpha})}\,\mathrm{d}x\) factors out \(\sqrt{E}\,x_{t}\propto E^{1/2+1/\alpha}\), and Bohr-Sommerfeld sets this \(\propto n\), giving \(E\propto n^{2\alpha/(\alpha+2)}\).)

For large \(n\) the spacing is the discrete derivative,

The relative spacing is therefore

So although the absolute spacing \(\Delta E\) may grow (\(\beta > 1\)), constant (\(\beta = 1\)), or shrink (\(\beta < 1\)), the fractional spacing always vanishes as \(1/n\). At high quantum numbers neighbouring levels are indistinguishable on the scale of the energy itself: the spectrum becomes quasi-continuous, exactly as a classical particle with continuously variable energy would have. This is the correspondence principle — quantum discreteness fades into the classical continuum as \(n\to\infty\).

(c) For hydrogen, \(E_{n} = -\text{Ry}/n^{2}\) with \(\text{Ry} = 13.6\,\text{eV}\). The spacing at large \(n\) is

The classical orbital frequency predicted by the correspondence principle is then

This \(n^{-3}\) scaling is exactly the Kepler result: the orbital radius grows as \(r_{n}\propto n^{2}\) (Problem 6a), and Kepler’s third law for a \(1/r\) potential gives a period \(T\propto r^{3/2}\propto n^{3}\), hence \(\omega_\text{cl} = 2\pi/T\propto n^{-3}\). A Rydberg electron in a high-\(n\) state orbits slowly, and when it drops to the adjacent level \(n \to n-1\) it radiates a photon of frequency \(\omega \approx \omega_\text{cl}\) — the quantum transition frequency between adjacent high-\(n\) levels coincides with the classical orbital frequency. This agreement between the quantum emission spectrum and classical electrodynamics is the original setting in which Bohr stated the correspondence principle.